C. Geometry and Algebra of Vectors · Vectors in Two-Space and Three-Space

Solving Geometry Problems with Vectors

C2.3 C2.2About 70 minutes

Learning goals

I can find the midpoint of a segment and the point dividing a segment in a given ratio, in two-space and three-space.
I can write vectors in a figure in terms of two non-collinear vectors and compare coefficients.
I can use vectors to prove geometric results such as the midsegment theorem and Varignon's theorem.
I can apply division-point ideas to real-world position problems.

Words to know

division point
A point P on segment AB with AP : PB = m : n; 𝐎𝐏 = (n𝐎𝐀 + m𝐎𝐁)/(m + n).

midsegment
A segment joining the midpoints of two sides of a triangle.

comparing coefficients
If 𝐮, 𝐯 are not collinear and a𝐮 + b𝐯 = c𝐮 + d𝐯, then a = c and b = d.

centroid
The point where the three medians of a triangle meet; 𝐎𝐆 = ⅓(𝐚 + 𝐛 + 𝐜).

Proofs without coordinates

In earlier grades you may have verified properties like "the diagonals of a parallelogram bisect each other" with coordinates (analytic geometry in Grade 10) or with congruent triangles. Vectors give a second, often shorter, way to prove them. They also give simple formulas for points that divide a segment in a given ratio, which you need for computer graphics, surveying and navigation ("the drone is three-fifths of the way to the depot"). This lesson puts all three operations (addition, subtraction and scalar multiplication) and their properties to work.

Two key tools

1. Midpoints and division points. If M is the midpoint of AB, then

𝐎𝐌 = 𝐎𝐀 + ½𝐀𝐁 = ½(𝐎𝐀 + 𝐎𝐁)

More generally, if P is on segment AB with AP : PB = m : n, then P is m/(m + n) of the way from A to B:

𝐎𝐏 = 𝐎𝐀 + (m/(m + n))𝐀𝐁 = (n𝐎𝐀 + m𝐎𝐁) / (m + n)

2. Comparing coefficients. If 𝐮 and 𝐯 are not collinear and a𝐮 + b𝐯 = c𝐮 + d𝐯, then a = c and b = d. (You proved why in the challenge of the lesson on subtraction and scalar multiples.) This lets you turn one vector equation into two ordinary equations.

Strategy for a vector proof: choose two non-collinear vectors (often two sides from one vertex), write every other vector in terms of them, then compare.

Worked example

Example 1: The midsegment theorem

In △ABC, M and N are the midpoints of AB and AC. Prove that MN is parallel to BC and half as long.

Let 𝐛 = 𝐀𝐁 and 𝐜 = 𝐀𝐂. Then 𝐀𝐌 = ½𝐛 and 𝐀𝐍 = ½𝐜.
𝐌𝐍 = 𝐌𝐀 + 𝐀𝐍 = −½𝐛 + ½𝐜 = ½(𝐜 − 𝐛).
𝐁𝐂 = 𝐁𝐀 + 𝐀𝐂 = −𝐛 + 𝐜 = 𝐜 − 𝐛.
So 𝐌𝐍 = ½𝐁𝐂. A positive scalar multiple means MN ∥ BC (same direction), and |𝐌𝐍| = ½|𝐁𝐂|. ∎

𝐌𝐍 = ½𝐁𝐂, so MN ∥ BC and MN = ½BC.

Writing everything in terms of 𝐛 = 𝐀𝐁 and 𝐜 = 𝐀𝐂 shows at once that 𝐌𝐍 is exactly half of 𝐁𝐂.

Worked example

Example 2: A point dividing a segment

Find the point P on segment AB with AP : PB = 1 : 2, for A(2, −1, 4) and B(8, 5, −2).

P is 1/(1 + 2) = 1/3 of the way from A to B.
𝐀𝐁 = [6, 6, −6], so (1/3)𝐀𝐁 = [2, 2, −2].
𝐎𝐏 = 𝐎𝐀 + [2, 2, −2] = [4, 1, 2].
Check with the formula: (2𝐎𝐀 + 1𝐎𝐁)/3 = ([4, −2, 8] + [8, 5, −2])/3 = [12, 3, 6]/3 = [4, 1, 2] ✓

P(4, 1, 2)

Splitting AB into 1 + 2 = 3 equal parts puts P one part from A and two parts from B.

Worked example

Example 3: Diagonals of a parallelogram bisect each other

Prove that the diagonals of any parallelogram bisect each other.

Let ABCD be a parallelogram with 𝐚 = 𝐀𝐁 and 𝐝 = 𝐀𝐃 (not collinear). Then 𝐀𝐂 = 𝐚 + 𝐝 and 𝐁𝐃 = 𝐝 − 𝐚.
Let the diagonals meet at E. E is on AC, so 𝐀𝐄 = s(𝐚 + 𝐝) for some scalar s. E is on BD, so 𝐁𝐄 = t(𝐝 − 𝐚) for some t.
Another path to E: 𝐀𝐄 = 𝐀𝐁 + 𝐁𝐄 = 𝐚 + t(𝐝 − 𝐚) = (1 − t)𝐚 + t𝐝.
Compare coefficients with s𝐚 + s𝐝: s = 1 − t and s = t. So t = ½ and s = ½.
s = ½ means E is the midpoint of AC; t = ½ means E is the midpoint of BD. ∎

Both diagonals are cut exactly in half at E.

Worked example

Example 4: Varignon's theorem

P, Q, R, S are the midpoints of sides AB, BC, CD, DA of any quadrilateral ABCD. Prove that PQRS is a parallelogram.

In △ABC, P and Q are midpoints of AB and BC, so by the midsegment theorem 𝐏𝐐 = ½𝐀𝐂.
In △ADC, S and R are midpoints of AD and DC, so 𝐒𝐑 = ½𝐀𝐂.
Therefore 𝐏𝐐 = 𝐒𝐑: one pair of opposite sides is parallel and equal in length.
A quadrilateral with one pair of opposite sides parallel and equal is a parallelogram. ∎

PQRS is always a parallelogram, even when ABCD is irregular.

Join the midpoints of the sides of any quadrilateral and you always get a parallelogram (shaded). Each of its sides is half of a diagonal (dashed).

Just for fun

Varignon's theorem works for any four-sided shape you can draw, however lopsided. Mathematicians love a theorem that can't be broken by bad drawing.

Hands-on activity

Midpoint magic

You need: two sheets of paper, a ruler, a pencil and (optional) coloured pencils.

Draw a large, lopsided quadrilateral ABCD (no two sides parallel).
Find the midpoint of each side by measuring with the ruler (or by folding the paper so the two endpoints meet). Label them P, Q, R, S in order.
Join P, Q, R, S. Measure all four sides of PQRS and compare opposite sides.
Draw the diagonals AC and BD of the big quadrilateral and measure them. Compare PQ with half of AC.
On the second sheet, draw a triangle, mark the point one third of the way along each median from the side's midpoint toward the opposite vertex, and see what happens.

What to look for: in step 3, opposite sides of PQRS should match to within a millimetre or two, and PQ should be about half of AC. In step 5, all three marked points should land on the same point, the centroid, which divides each median in the ratio 2 : 1 from the vertex (see the challenge).

Your turn

Find the midpoint of A(−3, 5) and B(7, −1).

Find the point P on segment AB with AP : PB = 2 : 1, for A(0, 0) and B(9, 12).

In △OAB, let 𝐚 = 𝐎𝐀 and 𝐛 = 𝐎𝐁, and let M be the midpoint of AB. Write 𝐎𝐌 in terms of 𝐚 and 𝐛.

In the same triangle, write 𝐀𝐁 and 𝐀𝐌 in terms of 𝐚 and 𝐛.

What two facts follow from 𝐌𝐍 = ½𝐁𝐂?

Find the point P dividing A(1, 2, 3) to B(7, −4, 9) in the ratio 1 : 2.

Show that A(1, 1), B(4, 3), C(6, 7) and D(3, 5) are the vertices of a parallelogram ABCD.

More practice: warm-up

Find the midpoint of (2, 4) and (6, 8).

Write the midpoint formula in vector form.

If AP : PB = 1 : 3, what fraction of the way from A to B is P?

If 𝐏𝐐 = 2𝐑𝐒, what can you say about PQ and RS?

𝐮 and 𝐯 are not collinear, and a𝐮 + b𝐯 = 3𝐮 + 5𝐯. Find a and b.

More practice: on your own

Find the point P on segment AB with AP : PB = 3 : 1, for A(2, 0, −1) and B(6, 8, 3).

In △ABC, D is on BC with BD : DC = 1 : 2. With 𝐛 = 𝐀𝐁 and 𝐜 = 𝐀𝐂, write 𝐀𝐃 in terms of 𝐛 and 𝐜.

A is (−1, 2) and B is (5, 5). Find the point C on line AB, beyond B, with BC = ½AB.

Check Varignon's theorem for A(0, 0), B(6, 0), C(8, 4), D(2, 6).

A drone flies in a straight line from depot D(0, 0, 0) to a farm F(600, 450, 0) (metres), then climbs. Where is it when it has covered 3/5 of the trip from D to F, and how far is it from F?

Challenge

In △ABC, let 𝐚, 𝐛, 𝐜 be the position vectors of A, B, C. Show that the point G with 𝐎𝐆 = ⅓(𝐚 + 𝐛 + 𝐜) lies on the median from A, two thirds of the way from A to the midpoint of BC.

Prove that if 𝐎𝐏 = s𝐎𝐀 + t𝐎𝐁 with s + t = 1, then P lies on line AB.

Find the two points that divide the segment from A(1, −2, 4) to B(10, 4, −2) into three equal parts.

Check your understanding

Quiz

What is the midpoint of (1, −3, 5) and (5, 7, −1)?

In △ABC, M and N are the midpoints of AB and AC. Which is true?

Which point P divides the segment from A(0, 0) to B(10, 5) so that AP : PB = 3 : 2?

If 𝐏𝐐 = −3𝐑𝐒, which statement is true?

M is the midpoint of AB. What is 𝐎𝐌?

𝐮 and 𝐯 are not collinear and a𝐮 + b𝐯 = 3𝐮 − 2𝐯. What are a and b?

For parents and teachers

Big idea (C2.3, C2.2): with two well-chosen vectors, many geometry facts reduce to a line or two of algebra. Division points are found by moving the right fraction of the way along a vector.

How to help: The "midpoint magic" activity is a nice surprise: any lopsided quadrilateral produces a parallelogram. Ask your teen to explain why using the midsegment theorem.

Common difficulties:

Using the ratio m : n as the fraction m/n instead of m/(m + n).
Comparing coefficients when the two vectors are collinear (not allowed).
Losing track of direction when writing a path (𝐌𝐀 = −𝐀𝐌).
Treating a vector proof as "checking a picture"; the proof must work for any shape.